17 VECTOR CALCULUS.

17 VECTOR CALCULUS

So far, we have considered special types of surfaces:
VECTOR CALCULUS So far, we have considered special types of surfaces: Cylinders Quadric surfaces Graphs of functions of two variables Level surfaces of functions of three variables

VECTOR CALCULUS Here, we use vector functions to describe more general surfaces, called parametric surfaces, and compute their areas.

VECTOR CALCULUS Then, we take the general surface area formula and see how it applies to special surfaces.

Parametric Surfaces and their Areas
VECTOR CALCULUS 17.6 Parametric Surfaces and their Areas In this section, we will learn about: Various types of parametric surfaces and computing their areas using vector functions.

INTRODUCTION We describe a space curve by a vector function r(t) of a single parameter t. Similarly, we can describe a surface by a vector function r(u, v) of two parameters u and v.

INTRODUCTION Equation 1 We suppose that r(u, v) = x(u, v) i + y(u, v) j + z (u, v) k is a vector-valued function defined on a region D in the uv-plane.

INTRODUCTION So x, y, and z—the component functions of r—are functions of the two variables u and v with domain D.

PARAMETRIC SURFACE Equations 2 The set of all points (x, y, z) in such that x = x(u, v) y = y(u, v) z = z(u, v) and (u, v) varies throughout D, is called a parametric surface S. Equations 2 are called parametric equations of S.

Each choice of u and v gives a point on S.
PARAMETRIC SURFACES Each choice of u and v gives a point on S. By making all choices, we get all of S.

PARAMETRIC SURFACES In other words, the surface S is traced out by the tip of the position vector r(u, v) as (u, v) moves throughout the region D. Fig , p. 1106

PARAMETRIC SURFACES Example 1 Identify and sketch the surface with vector equation r(u, v) = 2 cos u i + v j + 2 sin u k The parametric equations for this surface are: x = 2 cos u y = v z = 2 sin u

PARAMETRIC SURFACES Example 1 So, for any point (x, y, z) on the surface, we have: x2 + z2 = 4 cos2u + 4 sin2u = 4 This means that vertical cross-sections parallel to the xz-plane (that is, with y constant) are all circles with radius 2.

PARAMETRIC SURFACES Example 1 Since y = v and no restriction is placed on v, the surface is a circular cylinder with radius 2 whose axis is the y-axis. Fig , p. 1106

In Example 1, we placed no restrictions on the parameters u and v.
PARAMETRIC SURFACES In Example 1, we placed no restrictions on the parameters u and v. So, we obtained the entire cylinder.

PARAMETRIC SURFACES If, for instance, we restrict u and v by writing the parameter domain as ≤ u ≤ π/ ≤ v ≤ 3 then x ≥ 0 z ≥ 0 0 ≤ y ≤ 3

In that case, we get the quarter-cylinder with length 3.
PARAMETRIC SURFACES In that case, we get the quarter-cylinder with length 3. Fig , p. 1107

PARAMETRIC SURFACES If a parametric surface S is given by a vector function r(u, v), then there are two useful families of curves that lie on S—one with u constant and the other with v constant. These correspond to vertical and horizontal lines in the uv-plane.

PARAMETRIC SURFACES Keeping u constant by putting u = u0, r(u0, v) becomes a vector function of the single parameter v and defines a curve C1 lying on S. Fig , p. 1107

GRID CURVES Similarly, keeping v constant by putting v = v0, we get a curve C2 given by r(u, v0) that lies on S. We call these curves grid curves. Fig , p. 1107

In Example 1, for instance, the grid curves obtained by:
Letting u be constant are horizontal lines. Letting v be constant are circles.

GRID CURVES In fact, when a computer graphs a parametric surface, it usually depicts the surface by plotting these grid curves—as we see in the following example.

Use a computer algebra system to graph the surface
GRID CURVES Example 2 Use a computer algebra system to graph the surface r(u, v) = <(2 + sin v) cos u, (2 + sin v) sin u, u + cos v> Which grid curves have u constant? Which have v constant?

GRID CURVES Example 2 We graph the portion of the surface with parameter domain ≤ u ≤ 4π, 0 ≤ v ≤ 2π It has the appearance of a spiral tube. Fig , p. 1107

GRID CURVES Example 2 To identify the grid curves, we write the corresponding parametric equations: x = (2 + sin v) cos u y = (2 + sin v) sin u z = u + cos v

If v is constant, then sin v and cos v are constant.
GRID CURVES Example 2 If v is constant, then sin v and cos v are constant. So, the parametric equations resemble those of the helix in Example 4 in Section 14.1

So, the grid curves with v constant are the spiral curves.
Example 2 So, the grid curves with v constant are the spiral curves. We deduce that the grid curves with u constant must be the curves that look like circles. Fig , p. 1107

GRID CURVES Example 2 Further evidence for this assertion is that, if u is kept constant, u = u0, then the equation z = u0 + cos v shows that the z-values vary from u0 – 1 to u0 + 1.

PARAMETRIC REPRESENTATION
In Examples 1 and 2 we were given a vector equation and asked to graph the corresponding parametric surface. In the following examples, however, we are given the more challenging problem of finding a vector function to represent a given surface. In the rest of the chapter, we will often need to do exactly that.

PARAMETRIC REPRESENTATIONS
Example 3 Find a vector function that represents the plane that: Passes through the point P0 with position vector r0. Contains two nonparallel vectors a and b.

Example 3 If P is any point in the plane, we can get from P0 to P by moving a certain distance in the direction of a and another distance in the direction of b. So, there are scalars u and v such that: = ua + vb

Example 3 The figure illustrates how this works, by means of the Parallelogram Law, for the case where u and v are positive. See also Exercise 40 in Section 13.2 Fig , p. 1108

Example 3 If r is the position vector of P, then So, the vector equation of the plane can be written as: r(u, v) = r0 + ua + vb where u and v are real numbers.

Example 3 If we write r = <x, y, z> r0 = <x0, y0, z0> a = <a1, a2, a3> b = <b1, b2, b3> we can write the parametric equations of the plane through the point (x0, y0, z0) as: x = x0 + ua1 + vb y = y0 + ua2 + vb z = z0 + ua3 + vb3

Example 4 Find a parametric representation of the sphere x2 + y2 + z2 = a2 The sphere has a simple representation ρ = a in spherical coordinates. So, let’s choose the angles Φ and θ in spherical coordinates as the parameters (Section 15.8).

Example 4 Then, putting ρ = a in the equations for conversion from spherical to rectangular coordinates (Equations 1 in Section 15.8), we obtain: x = a sin Φ cos θ y = a sin Φ sin θ z = a cos Φ as the parametric equations of the sphere.

Example 4 The corresponding vector equation is: r(Φ, θ) = a sin Φ cos θ i + a sin Φ sin θ j + a cos Φ k We have 0 ≤ Φ ≤ π and 0 ≤ θ ≤ 2π. So, the parameter domain is the rectangle D = [0, π] x [0, 2π]

Example 4 The grid curves with: Φ constant are the circles of constant latitude (including the equator). θ constant are the meridians (semicircles), which connect the north and south poles.

APPLICATIONS—COMPUTER GRAPHICS
One of the uses of parametric surfaces is in computer graphics.

COMPUTER GRAPHICS The figure shows the result of trying to graph the sphere x2 + y2 + z2 = 1 by: Solving the equation for z. Graphing the top and bottom hemispheres separately. Fig , p. 1108

COMPUTER GRAPHICS Part of the sphere appears to be missing because of the rectangular grid system used by the computer. Fig , p. 1108

COMPUTER GRAPHICS The much better picture here was produced by a computer using the parametric equations found in Example 4. Fig , p. 1108

Example 5 Find a parametric representation for the cylinder x2 + y2 = ≤ z ≤ 1 The cylinder has a simple representation r = 2 in cylindrical coordinates. So, we choose as parameters θ and z in cylindrical coordinates.

Example 5 Then the parametric equations of the cylinder are x = 2 cos θ y = 2 sin θ z = z where: 0 ≤ θ ≤ 2π 0 ≤ z ≤ 1

Example 6 Find a vector function that represents the elliptic paraboloid z = x2 + 2y2 If we regard x and y as parameters, then the parametric equations are simply x = x y = y z = x2 + 2y2 and the vector equation is r(x, y) = x i + y j + (x2 + 2y2) k

In general, a surface given as the graph of a function of x and y—an equation of the form z = f(x, y)—can always be regarded as a parametric surface by: Taking x and y as parameters. Writing the parametric equations as x = x y = y z = f(x, y)

PARAMETRIZATIONS Parametric representations (also called parametrizations) of surfaces are not unique. The next example shows two ways to parametrize a cone.

PARAMETRIZATIONS Example 7 Find a parametric representation for the surface that is, the top half of the cone z2 = 4x2 + 4y2

PARAMETRIZATIONS E. g. 7—Solution 1 One possible representation is obtained by choosing x and y as parameters: x = x y = y So, the vector equation is:

PARAMETRIZATIONS E. g. 7—Solution 2 Another representation results from choosing as parameters the polar coordinates r and θ. A point (x, y, z) on the cone satisfies: x = r cos θ y = r sin θ

PARAMETRIZATIONS E. g. 7—Solution 2 So, a vector equation for the cone is r(r, θ) = r cos θ i + r sin θ j + 2r k where: r ≥ 0 0 ≤ θ ≤ 2π

In certain situations, though, Solution 2 might be preferable.
PARAMETRIZATIONS For some purposes, the parametric representations in Solutions 1 and 2 are equally good. In certain situations, though, Solution 2 might be preferable.

PARAMETRIZATIONS For instance, if we are interested only in the part of the cone that lies below the plane z = 1, all we have to do in Solution 2 is change the parameter domain to: ≤ r ≤ ½ 0 ≤ θ ≤ 2π

SURFACES OF REVOLUTION
Surfaces of revolution can be represented parametrically and thus graphed using a computer.

For instance, let’s consider the surface S obtained by rotating the curve y = f(x) a ≤ x ≤ b about the x-axis, where f(x) ≥ 0.

Let θ be the angle of rotation as shown. Fig , p. 1110

Equations 3 If (x, y, z) is a point on S, then x = x y = f(x) cos θ z = f(x) sin θ Fig , p. 1110

Thus, we take x and θ as parameters and regard Equations 3 as parametric equations of S. The parameter domain is given by: a ≤ x ≤ b 0 ≤ θ ≤ 2π

Example 8 Find parametric equations for the surface generated by rotating the curve y = sin x, 0 ≤ x ≤ 2π, about the x-axis. Use these equations to graph the surface of revolution.

Example 8 From Equations 3, The parametric equations are: x = x y = sin x cos θ z = sin x sin θ The parameter domain is: ≤ x ≤ 2π 0 ≤ θ ≤ 2π

Example 8 Using a computer to plot these equations and rotate the image, we obtain this graph. Fig , p. 1110

We can adapt Equations 3 to represent a surface obtained through revolution about the y- or z-axis. See Exercise 30.

TANGENT PLANES We now find the tangent plane to a parametric surface S traced out by a vector function r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k at a point P0 with position vector r(u0, v0).

TANGENT PLANES Keeping u constant by putting u = u0, r(u0, v) becomes a vector function of the single parameter v and defines a grid curve C1 lying on S. Fig , p. 1110

TANGENT PLANES Equation 4 The tangent vector to C1 at P0 is obtained by taking the partial derivative of r with respect to v:

TANGENT PLANES Similarly, keeping v constant by putting v = v0, we get a grid curve C2 given by r(u, v0) that lies on S. Fig , p. 1110

Its tangent vector at P0 is:
TANGENT PLANES Equation 5 Its tangent vector at P0 is:

SMOOTH SURFACE If ru x rv is not 0, then the surface is called smooth (it has no “corners”). For a smooth surface, the tangent plane is the plane that contains the tangent vectors ru and rv , and the vector ru x rv is a normal vector to the tangent plane.

TANGENT PLANES Example 9 Find the tangent plane to the surface with parametric equations x = u y = v z = u + 2v at the point (1, 1, 3).

We first compute the tangent vectors:
TANGENT PLANES Example 9 We first compute the tangent vectors:

Thus, a normal vector to the tangent plane is:
TANGENT PLANES Example 9 Thus, a normal vector to the tangent plane is:

TANGENT PLANES Example 9 Notice that the point (1, 1, 3) corresponds to the parameter values u = 1 and v = 1. So, the normal vector there is: –2 i + 4 j + 4 k

TANGENT PLANES Example 9 Therefore, an equation of the tangent plane at (1, 1, 3) is: –2(x – 1) – 4(y – 1) + 4(z – 3) = 0 or x + 2y – 2z + 3 = 0

TANGENT PLANES The figure shows the self-intersecting surface in Example 9 and its tangent plane at (1, 1, 3). Fig , p. 1111

SURFACE AREA Now, we define the surface area of a general parametric surface given by Equation 1.

SURFACE AREAS For simplicity, we start by considering a surface whose parameter domain D is a rectangle, and we divide it into subrectangles Rij.

Let’s choose (ui*, vj*) to be the lower left corner of Rij.
SURFACE AREAS Let’s choose (ui*, vj*) to be the lower left corner of Rij. Fig , p. 1111

PATCH The part Sij of the surface S that corresponds to Rij is called a patch and has the point Pij with position vector r(ui*, vj*) as one of its corners. Fig , p. 1111

SURFACE AREAS Let ru* = ru(ui*, vj*) and rv* = rv(ui*, vj*) be the tangent vectors at Pij as given by Equations 5 and 4.

SURFACE AREAS The figure shows how the two edges of the patch that meet at Pij can be approximated by vectors. Fig a, p. 1112

SURFACE AREAS These vectors, in turn, can be approximated by the vectors Δu ru* and Δv rv* because partial derivatives can be approximated by difference quotients. So, we approximate Sij by the parallelogram determined by the vectors Δu ru* and Δv rv*.

This parallelogram is shown here.
SURFACE AREAS This parallelogram is shown here. It lies in the tangent plane to S at Pij. Fig b, p. 1112

SURFACE AREAS The area of this parallelogram is: So, an approximation to the area of S is:

SURFACE AREAS Our intuition tells us that this approximation gets better as we increase the number of subrectangles. Also, we recognize the double sum as a Riemann sum for the double integral This motivates the following definition.

Suppose a smooth parametric surface S is:
SURFACE AREAS Definition 6 Suppose a smooth parametric surface S is: Given by r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k (u, v) D Covered just once as (u, v) ranges throughout the parameter domain D.

Then, the surface area of S is where:
SURFACE AREAS Definition 6 Then, the surface area of S is where:

Find the surface area of a sphere of radius a.
SURFACE AREAS Example 10 Find the surface area of a sphere of radius a. In Example 4, we found x = a sin Φ cos θ, y = a sin Φ sin θ, z = a cos Φ where the parameter domain is: D = {(Φ, θ) | 0 ≤ Φ ≤ π, 0 ≤ θ ≤ 2π)

We first compute the cross product of the tangent vectors:
SURFACE AREAS Example 10 We first compute the cross product of the tangent vectors:

SURFACE AREAS Example 10

SURFACE AREAS Example 10 Thus, since sin Φ ≥ 0 for 0 ≤ Φ ≤ π.

Hence, by Definition 6, the area of the sphere is:
SURFACE AREAS Example 10 Hence, by Definition 6, the area of the sphere is:

SURFACE AREA OF THE GRAPH OF A FUNCTION
Now, consider the special case of a surface S with equation z = f(x, y), where (x, y) lies in D and f has continuous partial derivatives. Here, we take x and y as parameters. The parametric equations are: x = x y = y z = f(x, y)

GRAPH OF A FUNCTION Equation 7 Thus, and

GRAPH OF A FUNCTION Equation 8 Thus, we have:

Then, the surface area formula in Definition 6 becomes:
GRAPH OF A FUNCTION Formula 9 Then, the surface area formula in Definition 6 becomes:

GRAPH OF A FUNCTION Example 11 Find the area of the part of the paraboloid z = x2 + y2 that lies under the plane z = 9. The plane intersects the paraboloid in the circle x2 + y2 = 9, z = 9

GRAPH OF A FUNCTION Example 11 Therefore, the given surface lies above the disk D with center the origin and radius 3. Fig , p. 1113

GRAPH OF A FUNCTION Example 11 Using Formula 9, we have:

Converting to polar coordinates, we obtain:
GRAPH OF A FUNCTION Example 11 Converting to polar coordinates, we obtain:

SURFACE AREA The question remains: Is our definition of surface area (Definition 6) consistent with the surface area formula from single-variable calculus (Formula 4 in Section 9.2)?

SURFACE AREA We consider the surface S obtained by rotating the curve y = f(x), a ≤ x ≤ b about the x-axis, where: f(x) ≥ 0. f’ is continuous.

From Equations 3, we know that parametric equations of S are:
SURFACE AREA From Equations 3, we know that parametric equations of S are: x = x y = f(x) cos θ z = f(x) sin θ a ≤ x ≤ b ≤ θ ≤ 2π

To compute the surface area of S, we need the tangent vectors

SURFACE AREA Thus,

SURFACE AREA Hence, because f(x) ≥ 0.

SURFACE AREA Thus, the area of S is:

SURFACE AREA This is precisely the formula that was used to define the area of a surface of revolution in single-variable calculus (Formula 4 in Section 9.2).

17 VECTOR CALCULUS.

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